The Robertson-Seymour theorem on graph minors leads to some interesting conundrums.

The theorem states that any minor-closed class of graphs can be described by a finite number of excluded minors. As testing for the presence of any given minor can be done in cubic time (albeit with astronomical constants) this implies that there exists a polynomial time algorithm for testing membership in any minor-closed class of graphs. Hence it seems reasonable that problem should be deemed to be in P.

However the RS theory does not give us even the faintest clue as to how to determine the guaranteed-finite set of excluded minors, and until we have these at hand, we may not have any algorithm of any sort. Worse still, there is no known algorithm to actually find the excluded minors and even if you have a big list of them, there is no way that I know to verify that the list is actually complete. In fact, could it perhaps actually be undecidable to find the list of excluded minors?

So, does it make sense to view a problem as being simultaneously polynomial-time and undecidable? It seems a bit odd to me (who is not a particular expert in complexity) but maybe it's quite routine?

Just a tip on asking your question: you really have three questions in your post, and it would have helped the quality of the answers if you had been more intentional about it, in particular by numbering them. Because some posters are answering 1 only, some 2 only, and no-one yet has addressed 3, and it makes it a mess to read when they don't announce at the outset which they're doing.
–
Thierry ZellDec 2 '10 at 13:48

6 Answers
6

Consider the following simplified example of the same phenomenon, which many students find clarifying.

Let $f(n)=1$, if there are $n$ consecutive $7$s in the decimal expansion of $\pi$, and otherwise $f(n)=0$. Is this function computable?

A naive attempt to compute $f(n)$ would simply proceed to search $\pi$ for $n$ consecutive $7$s. If found, the algorithm outputs $1$, but otherwise....and then the naive algorithm doesn't seem to know when to output $0$, and so students sometimes expect that $f$ is not computable.

But actually, $f$ is a computable function. If it happens that there are arbitrarily long sequences of $7$s in the decimal expansion of $\pi$, an open question, then $f$ is the constant $1$ function, which is certainly computable. Otherwise, there is some longest sequence of $7$s in $\pi$, having length $N$, and so $f$ is the function that is $1$ up to $N$ and then $0$ above $N$. And this function also is computable, for any particular $N$.

So the situation is that we have proved that $f$ is computable by exhibiting several algorithms, and proving that $f$ is definitely computed by one of them, but we don't know which one. (In fact, $f$ is linear time computable.) So we have proved that $f$ is a computable function, but by a pure existence proof that merely shows there is an algorithm computing $f$, without explicitly exhibiting it.

It seems to be the same phenomenon in your case, where you have a computable function, but you don't know which algorithm computes it.

Addition. Let me try to address Thierry Zell's concern about the
third question. To my way of thinking, the phenomenon of
the question is an instance of the problem of uniformity
of algorithms, a pervasively considered issue in
computability theory.

To illustrate, consider the question of whether a given
program $p$ halts on input $0$ before another program $q$.
Let $f_p(q)=1$ if it does and otherwise $f_p(q)=0$. Every
such function $f_p$ is computable, for similar reasons to
my $\pi$ function $f$ above, since either $p$ doesn't halt
at all on input $0$, in which case $f_p$ is identically
$0$, or $p$ does halt in $N$ steps, in which case we need
only run $q$ for $N$ steps to see if it halts, and give our
output for $f_p(q)$ by that time. So each $f_p$ is a
computable function. But the joint function
$f(p,q)=f_p(q)$, a binary function, is not computable (if
it were, then we could solve the halting problem: to decide
if $p$ halts on input $0$, design a program $q$ that would
take one step extra after a halt, and ask if $p$ halts
before $q$).

In other words, the function $f(p,q)$ is computable for any
fixed $p$, but not uniformly in $p$. And such uniformity
issues are ubiquitous in computability theory.

In the example of the question, each class of graphs is
decidable, but not uniformly so, since by Tony's answer
there is no uniform algorithm, given a description of the
class, to find the collection of excluded minors. But for
any such fixed class, the membership question is decidable.

The issue of whether a given algorithm is uniform in a
given parameter is a very common concern in computability
theory, and occurs throughout the subject.

Frank, I had the interpretation that if you have fifteen consecutive $7$s, then the first eleven of them would count as eleven consecutive $7$s. So the function really would be as I describe it. Otherwise, I suppose, one should speak of *maximal* consecutive sequences of $7$s, and I believe that it is an open question whether the corresponding function is computable. (Although if $\pi$ is normal, then this function also would be identically $1$.)
–
Joel David HamkinsDec 2 '10 at 17:03

This reminds me of a very similar problem on a problem set I once did regarding the definition of decidability of a language. The problem was this: Let L = {0} if there is life in the universe outside our solar system, and L = {1} if the only life in the universe is inside our solar system. Then the question is: is L decidable?
–
Taylor SuttonDec 3 '10 at 4:10

@Joel, a link to this fine answer has been added to the TCS StackExchange question "Do the undecidable attributes of P pose an obstruction to deciding P versus NP?" ... I wish to thank Alex ten Brink for drawing attention to this answer.
–
John SidlesJun 28 '11 at 13:13

As others have mentioned, the answer to your title question is strictly speaking no. With regards to your other questions, it has been proven that it is undecidable to compute the excluded minors for a minor-closed class $\mathcal{C}$, unless $\mathcal{C}$ is presented to you in a silly way. Of course, there is no paradox, because this does not imply that the related problem of determining if an input graph $G$ is in $\mathcal{C}$ is undecidable. Indeed, as you mention by the Robertson-Seymour theory, this second problem is not only decidable, but is in P.

I guess I should quantify what I mean by non-silly representations of minor-closed families. Fellows and Langston proved that
if your minor-closed class $\mathcal{C}$ is given by a Turing machine $M$, then it is undecidable to compute an excluded minor characterization of $\mathcal{C}$. Courcelle, Downey, and Fellows proved that if $\mathcal{C}$ is instead given by a monadic second-order logic formula $\phi$, then it is also undecidable to compute an excluded minor characterization of $\mathcal{C}$.

There are positive results for certain minor-closed families. For example, this paper by Adler, Grohe, and Kreutzer shows that for any fixed $k$, they can compute the excluded minors for the class of graphs of tree-width at most $k$. For the undecidability results that I mentioned above, the references are:

M.R. Fellows and M.A. Langston. On search, decision and the efficiency of polynomial-time algorithms (extended abstract). In Proceedings of the 21st ACM Symposium on Theory of Computing, pages 501–512, 1989.

B. Courcelle, R.G. Downey, and M.R. Fellows. A note on the computability of graph minor obstruction sets for monadic second order ideals. Journal of Universal Computer Science, 3:1194–1198, 1997.

Of course, every problem in P is decidable by definition of P. This was mentioned in the previous answers.

But there is another problem here that hasn't been addressed yet:
you apparently are looking for an algorithm that takes as input a class closed under minors and a finite graph and decides whether or not the finite graph is in the class.
Or you are looking for an algorithm that takes a class closed under minors and produces an polynomial time algorithm that decides membership to the class.

Here is the problem: How do you present a class of graphs closed
under minors? A priori, it is not clear that every class of graphs that is closed under minors (usually a class containing graphs of infinitely many isomorphism classes) has a reasonable representation as a finite object (that can be treated algorithmically) at all.
By finite representation I mean a formula that defines the class in one way or other or something similar.

Now, the graph minor theorem gives us a nice representation of every such class: Just list
the finite set of forbidden minors. If this is your representation of the class, then you
get your polynomial time algorithm that decides membership for the class.

If you settle on another representation (and you have to come up with some uniform way to describe your class by finite objects to be able to say anything algorithmically at all, I would think), it might not be possible to come up with an algorithm that computes the finitely many forbidden minors from the representation of the class.

For any given minor-closed class of graphs, there is some finite set of excluded minors, and hence there is a polynomial time algorithm for testing membership of that class (we don't need to explicity know what the algorithm is, we simply know it exists.)

However, on the level above, writing out that algorithm explicitly involves finding a finite set of excluded minors explicitly, and that might be hard/undecidable.

More to the point, the 'level above' isn't even well-defined algorithmically - how are you specifying a 'minor-closed class' as input? A fairly general way would be to specify it as a first-order predicate perhaps, and then you are likely to run into undecidability problems. Of course if you make use of R-S results and specify the minor-closed class by a finite set of excluded minors, then the 'level above' becomes rather trivial.
–
ndkrempelDec 2 '10 at 9:36

I can't seem to find the paper now, but I recall reading a proof that the "level above" problem (given, I believe, a first-order predicate for the minor-closed class) is not decidable.
–
Henry TowsnerDec 2 '10 at 18:42

You just want "recursive" there instead of "recursively enumerable". Plenty of things are undecidable yet r.e. The theorems of PA for instance (and more to the point, any non-recursive r.e. set).
–
Ed DeanDec 2 '10 at 9:21

Indeed ! Should not be confused between "recursive" and "recursively enumerable". Thanks.
–
LamineDec 2 '10 at 9:36

6

This is strictly speaking true, but not really getting at the paradox that's bothering gordon-royle. One reasonable interpretation of Robertson-Seymour is that in some abstract non-constructive sense it proves the existence of a polynomial-time algorithm for a problem. To use the algorithm, you need a finite amount of data, but it's known that there's no algorithm for finding that data (Tony Huynh gives references in his answer). It's a pretty weird situation.
–
arsmathDec 2 '10 at 15:04

Is there some confusion between decision problems and the computation of their solutions ? A (decision) problem on the existence of a solution of length k can be in P but compute this solution can be exponential, nay currently impossible. This is precisely due to non-constructive proof procedure.
–
LamineDec 3 '10 at 8:28

It will be solved by either 2048 or
4096. I am currently somewhat pessimistic. The outcome will be the
truly worst case scenario: namely that
someone will prove “P=NP because there
are only finitely many obstructions to
the opposite hypothesis”; hence there
will exists a polynomial time solution
to SAT but we will never know its
complexity!